Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems } 1-44, \text { find } D_{x} y \text { using the rules of this section. } $$ $$ y=\left(x^{4}-1\right)\left(x^{2}+1\right) $$

Step-by-Step Solution

Answer

The derivative is \( 6x^5 + 4x^3 - 2x \).

Step 1: Understand the Problem

The problem asks us to find the derivative of the function \( y = (x^4 - 1)(x^2 + 1) \) with respect to \( x \). We will use the product rule to differentiate since the function is a product of two smaller functions.

Step 2: Recall the Product Rule

The product rule for differentiation states that if \( y = u(x)v(x) \), then \( \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \). Here, \( u(x) = x^4 - 1 \) and \( v(x) = x^2 + 1 \).

Step 3: Differentiate \( u(x) \)

Differentiate \( u(x) = x^4 - 1 \) to find \( u'(x) \). Using the power rule, \( u'(x) = 4x^3 \).

Step 4: Differentiate \( v(x) \)

Differentiate \( v(x) = x^2 + 1 \) to find \( v'(x) \). Using the power rule, \( v'(x) = 2x \).

Step 5: Apply the Product Rule

Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the product rule formula: \( D_{x}y = u'(x)v(x) + u(x)v'(x) = 4x^3(x^2 + 1) + (x^4 - 1)(2x) \).

Step 6: Simplify the Derivative Expression

Expand and simplify the terms: \( 4x^3(x^2 + 1) = 4x^5 + 4x^3 \) and \( (x^4 - 1)(2x) = 2x^5 - 2x \). Thus, \( D_{x}y = 4x^5 + 4x^3 + 2x^5 - 2x = 6x^5 + 4x^3 - 2x \).

Step 7: Write the Final Derivative

The derivative \( D_{x}y \) is \( 6x^5 + 4x^3 - 2x \). This is the final simplified expression of the derivative of the given function.

Key Concepts

Understanding the Product Rule

To differentiate functions that are products of other functions, we use the product rule. This rule simplifies the process of finding the derivative when you have two functions multiplied together, say \( u(x) \) and \( v(x) \). According to the product rule, the derivative \( \frac{dy}{dx} \) of \( y = u(x) \times v(x) \) is given by:

• \( u'(x)v(x) + u(x)v'(x) \)

Let's break it down:- \( u'(x) \) is the derivative of the first function \( u(x) \).- \( v(x) \) is the second function left as is.- \( u(x) \) is the first function left as is.- \( v'(x) \) is the derivative of the second function \( v(x) \).You take the derivative of the first function, multiply by the second function, and then add it to the first function multiplied by the derivative of the second function. This ensures you capture all the necessary parts of changes in both functions.

Exploring the Power Rule

The power rule is an essential tool in calculus differentiation. It helps simplify the process of finding derivatives of functions where the variable has an exponent. If you have a function in the form of \( x^n \), the power rule states that its derivative is:

• \( nx^{n-1} \)

Here's how it works:- Multiply the current exponent \( n \) by the function.- Decrease the exponent by one.
For example, if \( u(x) = x^4 - 1 \), the derivative \( u'(x) \) would be calculated by:- Finding \( 4x^{4-1} = 4x^3 \).Similarly, for \( v(x) = x^2 + 1 \):- The derivative \( v'(x) \) becomes \( 2x^{2-1} = 2x \).
Using the power rule not only simplifies the differentiation process but also prepares you to handle more complex expressions when combined with other rules like the product rule.

Simplifying Derivatives

Once we apply the product rule and power rule, the next step is to simplify the resultant expression. Simplification involves combining like terms and reducing the equation to its most concise form. In our original problem, we had to apply both rules:

• The product was expanded: \( 4x^3(x^2 + 1) + (x^4 - 1)(2x) \)
• This gives us: \( 4x^5 + 4x^3 + 2x^5 - 2x \)

To simplify:- Combine like terms, which are terms with the same power of \( x \).- Therefore, \( 4x^5 + 2x^5 = 6x^5 \) and \( 4x^3 \), \(- 2x \) remain as they are.
Thus, the simplified form of the derivative is \( 6x^5 + 4x^3 - 2x \). Simplification helps in understanding the behavior of the function's rate of change and is critical for solving real-world problems.